C语言 将节点与左节点进行比较后删除节点

Question

所以我在删除节点时编写这个程序，如果它大于它左侧的节点，并找到迭代次数，之后没有节点被删除。我想出了这个，但int days始终保持在 0。

#include<stdio.h>
#include<malloc.h>
struct plants{
int val;
struct plants *next;
};
void printlist();
int main(){
int counter=0;
int size=0;
int days=0;
printf("Enter the number of plantsn");
scanf("%d",&size);
printf("Enter the amount of pesticide each plant has.n");
struct plants* head = NULL;
while( counter < size )
{
    struct plants * current = malloc(sizeof(struct plants)); 
    scanf( "%d", &current->val);                             
    current->next = head;                                    
    head = current;                                          
    counter ++;                                              
}
struct plants *temp = head;
printf("You have entered.n");
while(temp!=NULL){
    printf("%dt",temp->val);
    temp=temp->next;
}
struct plants *now = head;
while(counter<size){
    if(now->val < now->next->val){
        struct plants* nextNext = now->next->next;
        days++;
        free(now->next);
        now->next= nextNext;
        counter++;
    }
    else{
        now = now->next;
    }
}
printf("The days after which the plants stop dying %d.n",days);
}

Answer 1

你需要一个指向要删除的节点的指针的指针，这样你就可以把节点的后继节点放到它的位置：

int days = 0;
struct plants **now = &head;
while( *now != NULL && (*now)->next != NULL ) // do as long as there are two nodes to compare
{
    struct plants *next = (*now)->next; // successor of the node
    if ( (*now)->val < next->val )      // test if successor node is greater than node
    {
        free( *now );                   // free the node
        *now = next;                    // put successor of the node in place of the node
    }
    else
    {
        now = &((*now)->next);          // step one forward
        days ++;                        // increment counter because no node was deleted
    }
}

另一种解决方案是记住当前节点的前置节点：

int days = 0;
struct plants *now = head;  // start at head of list
struct plants *prev = NULL; // predecessor of head is NULL
while( now != NULL && now->next != NULL ) // do as long as there are two nodes to compare
{
    struct plants *next = now->next;  // successor of the node
    if ( now->val < next->val )       // test if successor node is greater than node
    {
        free( now );                   // free the node
        if ( prev == NULL )            // put successor of the node in place of the node
            head = next;               // the first node of the list was deleted
        else
            prev->next = next;         // successor of predecessor is predecessor of deleted node
        now = next;                    // step one forward
                                       // note "prev" does not change in this case
    }
    else
    {
        prev = now;
        now = now->next;               // step one forward
        days ++;                       // increment counter because no node was deleted
    }
}

Answer 2

看：

while( counter < size )
{
    struct plants * current = malloc(sizeof(struct plants)); 
    scanf( "%d", &current->val);                             
    current->next = head;                                    
    head = current;                                          
    counter ++;                                              
}
//counter == size
struct plants *temp = head;
printf("You have entered.n");
while(temp!=NULL){
    printf("%dt",temp->val);
    temp=temp->next;
}
struct plants *now = head;
//counter == size
while(counter<size){
    if(now->val < now->next->val){
        struct plants* nextNext = now->next->next;
        days++;
        free(now->next);
        now->next= nextNext;
        counter++;
    }
    else{
        now = now->next;
    }
}

您没有重置counter的值